Cal 2 Final Questions

Question 1

How to solve this problem ?

Answer 1

Start doing the root convergence test since there is a kth power.

Then set the 3x + 4 between the interval (-1,1), and solve for x

-1 < 3x+4 < 1 → -5/3 < x < -1

Then find the average between these two values: (-5/3) - (-1) / 2

Question 2

Compute the slope at t = 0 of the curve that is given parametrically by y = sec(t) and x = tan(t) + 1

Answer 2

We start by knowing that the slope is known by function:

dy/dx = [dy/dt]/[dx/dt]

so we can take the derivative of both sides and solve at 0:

sec(t)tan(t) / sec²(t)

sin(0) = 0

Question 3

Explain how to solve

Answer 3

The remainder will always be Rₙ(x) = ⨍ⁿ⁺¹(c)(x - c)ⁿ⁺¹/(n +1)!

Where x is the approximation value, and c is a near estimation value like;y given in the problem:

⨍(x) = ln(x)
⨍’(x) = 1/x
⨍’‘(x) = -1/x²
⨍⁽³⁾(x) = 2/x³

so: R₂(0.9) = ⨍⁽³⁾(c) ⋅ (0.9 - 1)³ / 3!

R₂(0.9) = [2c⁻³/3⋅2] ⋅ (-1/10)³

R₂(0.9) = -1/3000c³

Question 4

Explain how to solve a washer method around an arbitrary axis

Answer 4

We use the typical Washer Method formula ∫π(R² - r²)dx

We take R to be the distance between our arbitrary axis and furthest function
R = (6 - g(x))

And we take r to be the distance between our arbitrary axis and the nearest function
r = (6 - f(x))

and solve our integral as: ∫π [(6 - g(x))² - (6 - f(x))²]dx